图中显示了 $y = f(x)$ 的图像，其渐近线为 $y = k$ 和 $x = 0$，且图像与 x 轴相交于 $(a,0)$。 在不同的图上，画出下列函数的图像，并给出所有渐近线的方程以及曲线与 x 轴和 y 轴交点的坐标。 画出 $y = \dfrac{1}{f(x)}$ 的图像。 画出 $y = \sqrt{f(-x)}$ 的图像。 曲线 $y = g(x)$ 依次经历变换 A、B 和 C： A. 以比例因子 2 沿平行于 x 轴的方向拉伸， B. 关于 y 轴反射，以及 C. 沿 y 轴方向平移 1 个单位。 若最终得到的曲线方程为 $y = 1 - \dfrac{1}{x}$，求 $g(x)$ 的表达式。

1. Determine the horizontal asymptote of the new graph.

The original graph has a horizontal asymptote $y = k$, which means as $x \to \pm\infty$, $f(x) \to k$.
Therefore, as $x \to \pm\infty$, $\dfrac{1}{f(x)} \to \dfrac{1}{k}$.
The new horizontal asymptote is $y = \dfrac{1}{k}$.

2. Determine the vertical asymptote of the new graph.

The vertical asymptote of $y = \dfrac{1}{f(x)}$ occurs where the denominator is zero, i.e., $f(x) = 0$.
The original graph cuts the x-axis at $x = a$, so $f(a) = 0$.
Therefore, the new vertical asymptote is $x = a$.

3. Determine the axial intercepts of the new graph.

The original graph has a vertical asymptote at $x = 0$, meaning $f(x) \to \pm\infty$ as $x \to 0$.
Therefore, $\dfrac{1}{f(x)} \to 0$ as $x \to 0$.
This means the new curve intersects the axes at the origin $(0,0)$.

1. Apply the reflection transformation $f(-x)$ to the key features of the original graph.

The graph of $y = f(-x)$ is a reflection of the graph of $y = f(x)$ in the $y$-axis.

• The original $x$-intercept $(a,0)$ reflects to $(-a,0)$.
• The vertical asymptote $x = 0$ (the $y$-axis) remains unchanged.
• The horizontal asymptote $y = k$ remains unchanged.
• Assuming the standard shape of a hyperbola where $f(x)$ approaches $k>0$ and cuts at $(a,0)$, $f(x) \ge 0$ for $x < 0$ and $x \ge a$. Reflecting this, $f(-x) \ge 0$ for $x > 0$ and $x \le -a$.

2. Determine the effect of the square root on the asymptotes and intercepts.

Taking the square root for $y = \sqrt{f(-x)}$ restricts the domain to regions where $f(-x) \ge 0$, which are $x \le -a$ and $x > 0$.

• Intercepts: The $x$-intercept at $(-a,0)$ remains $(-a,0)$ because $\sqrt{0}=0$. The curve will have an infinite (vertical) gradient at this root.
• Asymptotes: The horizontal asymptote $y = k$ becomes $y = \sqrt{k}$. The vertical asymptote remains $x = 0$ for the positive branch where $y \to \infty$ as $x \to 0^+$.

3. Deduce the final shape of the two branches for the sketch.

• Left branch ($x \le -a$): Starts at $(-a, 0)$ with a vertical tangent. As $x \to -\infty$, the $y$-values increase, asymptotically approaching the horizontal line $y = \sqrt{k}$ from below.
• Right branch ($x > 0$): Located entirely in the first quadrant. As $x \to 0^+$, $y \to \infty$. It decreases as $x \to \infty$, asymptotically approaching $y = \sqrt{k}$ from above.

1. Express the function after the three successive transformations in terms of $g(x)$.

Let the original curve be $y = g(x)$.
After transformation A (stretch parallel to the $x$-axis with scale factor 2):
$y = g\left(\frac{x}{2}\right)$

After transformation B (reflection in the $y$-axis, replacing $x$ with $-x$):
$y = g\left(-\frac{x}{2}\right)$

After transformation C (translation of 1 unit in the $y$-direction):
$y = g\left(-\frac{x}{2}\right) + 1$

2. Equate the transformed expression to the given final equation.

The resulting curve is given as $y = 1 - \frac{1}{x}$. Equating the two expressions for $y$:
$g\left(-\frac{x}{2}\right) + 1 = 1 - \frac{1}{x}$
$g\left(-\frac{x}{2}\right) = -\frac{1}{x}$

3. Solve for the original function $g(x)$ using an algebraic substitution.

Let $u = -\frac{x}{2}$, which implies $x = -2u$.
Substitute $x = -2u$ into the equation:
$g(u) = -\frac{1}{-2u}$
$g(u) = \frac{1}{2u}$
Replacing $u$ with $x$ gives the final expression:
$g(x) = \frac{1}{2x}$